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For simplicity, assume everythings occur on a smooth projective variaty $X$.

Dual bundle of the given line bundle $\mathcal L$ is determined by $\mathcal L$ and $c_1(\mathcal L)$.

$\mathcal L^*= \mathcal L (-2c_1(\mathcal L)) $

My question is, is their any similar relation between a vector bundle of rank >1 and its dual? Can dual vector bundle be described as a combination of original bundle and its data(e.g chern classes)? Even for a rank 2 case, I have no idea about this problem.

If you have one, please give me some short proof or sketch. Good reference is also very preferable. I appreciate any help.

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    $\begingroup$ What exactly do you mean by that formula for $\mathcal{L}^∗$? The chern class $c_1(\mathcal{L}^∗)\in H^2(X,\mathbb{Z})$ does not canonically determine a line bundle.. $\endgroup$
    – J.C. Ottem
    Apr 26, 2012 at 2:08
  • $\begingroup$ Algebro-geometric definition and differential geometric definition is slightly different. In algebraic geometry, Chern class lives in the Chow ring, which is the space of cycles modulo rational equivalance. For the line bundle, the divisor of a secrion is the first chern class. Lelong formula says its cohomology class recovers the cohomologically defined chern class, so they are closely related. $\endgroup$
    – Choa
    Apr 26, 2012 at 2:22
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    $\begingroup$ @Choa: Essentially the statement you are making is that $\mathcal{O}(D)^*=\mathcal{O}(-D)=\mathcal{O}(D)\otimes \mathcal{O}(-2D)$, which while true, is hardly a numerical characterization of the dual. Expecting anything less tautological in the case of a more general vector bundle is unrealistic. $\endgroup$
    – Daniel Litt
    Apr 26, 2012 at 2:45
  • $\begingroup$ @Daniel: Your answer somehow clearifies my unrealistic thought...... Your right. $\endgroup$
    – Choa
    Apr 26, 2012 at 3:04

1 Answer 1

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If $E$ is locally free of rank 2 then $E^* \cong E(-c_1(E))$. The isomorphism is induced by the nondegenerate pairing $E\otimes E \to \Lambda^2E \cong O(c_1(E))$. For higher rank one can check that in general $E^*$ is not a twist of $E$.

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    $\begingroup$ While your negative statement is true, it is worth generalizing to $E^*\cong (\bigwedge^{n-1} E)(-c_1E)$. $\endgroup$
    – Ben Wieland
    Apr 26, 2012 at 3:44
  • $\begingroup$ @Ben Wieland: Of course, and the reason is the same. $\endgroup$
    – Sasha
    Apr 26, 2012 at 5:57
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    $\begingroup$ Oh my! It is just a special case of koszul complex and its self duallity! I have to study this topic more deeply. $\endgroup$
    – Choa
    Apr 26, 2012 at 6:05

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